If the base of a triangle and the ratio of the lengths of the other two unequal sides are given, then the vertex lies on

Options

A. straight line
B. circle
C. ellipse
D. parabola

Answer

B

Explanation

Let the base of the triangle be the line segment joining B(0, 0) and C(a, 0) and let the vertex be A(h, k), where a is fixed. Also, let
`(AB)/(AC)=lambda, lambda !=1`
`rArr AB^(2)=lambda^(2)AC^(2)`
`rArrh^(2)+k^(2)=lambda^(2){(h-a)^(2)+k^(2)}`
`rArr h^(2)(1-lambda^(2))+k^(2)(1-lambda^(2))+2a lambda^(2)h-a^(2)lambda^(2)=0`
So, A(h, k) lies on
`x^(2)+y^(2)+2a(lambda^(2))/(1-lambda^(2))x-(a^(2)lambda^(2))/(1-lambda^(2))=0`, which is a circle.